The solubility product of `BaCl_2` is `4.0 xx 10 ^(–8)` what will be its molar solubility in mol `dm ^-3`

Answer

To find the molar solubility of \( \text{BaCl}_2 \) given its solubility product \( K_{sp} = 4.0 \times 10^{-8} \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of barium chloride in water can be represented as:
\[ \text{BaCl}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define Molar Solubility
Let the molar solubility of \( \text{BaCl}_2 \) be \( s \) mol/dm³. When \( \text{BaCl}_2 \) dissolves, it produces:
- \( s \) mol/dm³ of \( \text{Ba}^{2+} \)
- \( 2s \) mol/dm³ of \( \text{Cl}^- \)

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{BaCl}_2 \) can be expressed as:
\[ K_{sp} = [\text{Ba}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (s)(2s)^2 \]
This simplifies to:
\[ K_{sp} = s \cdot 4s^2 = 4s^3 \]

### Step 4: Substitute the Value of \( K_{sp} \)
Now, we can substitute the given value of \( K_{sp} \):
\[ 4s^3 = 4.0 \times 10^{-8} \]

### Step 5: Solve for \( s \)
Dividing both sides by 4:
\[ s^3 = 1.0 \times 10^{-8} \]
Now, take the cube root of both sides:
\[ s = \sqrt[3]{1.0 \times 10^{-8}} = 1.0 \times 10^{-8/3} \approx 2.15 \times 10^{-3} \, \text{mol/dm}^3 \]

### Final Answer
The molar solubility of \( \text{BaCl}_2 \) is approximately \( 2.15 \times 10^{-3} \, \text{mol/dm}^3 \).

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